If $7,300 is invested at an interest rate of 10% per year, find the value of the investment at the end of 5 years if interest is compounded annually (once a year), semiannually (twice a year), monthly (12 times a year), daily (assume 365 days a year), or continuously. Round to the nearest cent. For each, use the correct compound interest formula from the following. A = Pert nt A = P(1+7) "² (a) Annual: (b) Semiannual: (c) Monthly: (d) Daily: (e) Continuously: or

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### Compound Interest Calculation

If $7,300 is invested at an interest rate of 10% per year, find the value of the investment at the end of 5 years if interest is compounded annually (once a year), semiannually (twice a year), monthly (12 times a year), daily (assume 365 days a year), or continuously. Round to the nearest cent. For each, use the correct compound interest formula from the following:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
\[ \text{or} \]
\[ A = Pe^{rt} \]

#### Let's breakdown the components:
- \( P \) is the principal amount (initial investment): $7,300
- \( r \) is the annual interest rate (decimal): 10% = 0.10
- \( t \) is the time in years: 5 years
- \( n \) is the number of times interest is compounded per year

#### Given Information to be calculated:
- (a) Annual:
- (b) Semiannual:
- (c) Monthly:
- (d) Daily:
- (e) Continuously:

The formulas and values calculated for each scenario will be:

1. **Annually** (\( n = 1 \)):
   \[ A = 7300 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 5} \]

2. **Semiannually** (\( n = 2 \)):
   \[ A = 7300 \left(1 + \frac{0.10}{2}\right)^{2 \cdot 5} \]

3. **Monthly** (\( n = 12 \)):
   \[ A = 7300 \left(1 + \frac{0.10}{12}\right)^{12 \cdot 5} \]

4. **Daily** (\( n = 365 \)):
   \[ A = 7300 \left(1 + \frac{0.10}{365}\right)^{365 \cdot 5} \]

5. **Continuously**:
   \[ A = 7300e^{0.10 \cdot 5} \]

The image does not contain any graphs or diagrams, promptly the mathematical formulas and the placeholders for the calculations
Transcribed Image Text:### Compound Interest Calculation If $7,300 is invested at an interest rate of 10% per year, find the value of the investment at the end of 5 years if interest is compounded annually (once a year), semiannually (twice a year), monthly (12 times a year), daily (assume 365 days a year), or continuously. Round to the nearest cent. For each, use the correct compound interest formula from the following: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] \[ \text{or} \] \[ A = Pe^{rt} \] #### Let's breakdown the components: - \( P \) is the principal amount (initial investment): $7,300 - \( r \) is the annual interest rate (decimal): 10% = 0.10 - \( t \) is the time in years: 5 years - \( n \) is the number of times interest is compounded per year #### Given Information to be calculated: - (a) Annual: - (b) Semiannual: - (c) Monthly: - (d) Daily: - (e) Continuously: The formulas and values calculated for each scenario will be: 1. **Annually** (\( n = 1 \)): \[ A = 7300 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 5} \] 2. **Semiannually** (\( n = 2 \)): \[ A = 7300 \left(1 + \frac{0.10}{2}\right)^{2 \cdot 5} \] 3. **Monthly** (\( n = 12 \)): \[ A = 7300 \left(1 + \frac{0.10}{12}\right)^{12 \cdot 5} \] 4. **Daily** (\( n = 365 \)): \[ A = 7300 \left(1 + \frac{0.10}{365}\right)^{365 \cdot 5} \] 5. **Continuously**: \[ A = 7300e^{0.10 \cdot 5} \] The image does not contain any graphs or diagrams, promptly the mathematical formulas and the placeholders for the calculations
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